(If any are cut off answer what you can see)
3A)

3b)

3C)

3D)

a)
| observed frequencey, O | expected proportion | expected frequency,E | (O-E) | (O-E)²/E | |
| 17 | 0.250 | 26.25 | -9.25 | 3.260 | |
| 57 | 0.400 | 42.00 | 15.00 | 5.357 | |
| 8 | 0.100 | 10.50 | -2.50 | 0.595 | |
| 23 | 0.250 | 26.25 | -3.25 | 0.402 |
chi square test statistic,X² = Σ(O-E)²/E =
9.614
level of significance, α= 0.01
Degree of freedom=k-1= 4 -
1 = 3
Critical value = 6.251 [ Excel
function: =chisq.inv.rt(α,df) ]
reject Ho
b)
| observed frequencey, O | expected proportion | expected frequency,E | (O-E) | (O-E)²/E | |
| 35 | 0.200 | 29.60 | 5.40 | 0.985 | |
| 35 | 0.200 | 29.60 | 5.40 | 0.985 | |
| 47 | 0.300 | 44.40 | 2.60 | 0.152 | |
| 31 | 0.300 | 44.40 | -13.40 | 4.044 |
chi square test statistic,X² = Σ(O-E)²/E =
6.167
level of significance, α= 0.05
Degree of freedom=k-1= 4 -
1 = 3
Do not reject Ho
Critical value = 7.815 [
Excel function: =chisq.inv.rt(α,df) ]
Please let me know in case of any doubt.
Thanks in advance!
Please upvote!
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